geometric distribution

Suppose that a random experiment has two possible outcomes, success with probability $p$ and failure with probability $q=1-p$. The experiment is repeated until a success happens. The number of trials before the success is a random variable $X$ with density function

$$f(x)=q^{(x-1)}p.$$

The distribution function determined by $f(x)$ is called a geometric distribution with parameter $p$ and it is given by

$$F(x)=\sum _{k\le x}{q}^{(k-1)}p.$$

The picture shows the graph for $f(x)$ with $p=1/4$. Notice the quick decreasing. An interpretation is that a long run of failures is very unlikely.

We can use the moment generating function method in order to get the mean and variance. This function is

$$G(t)=\sum _{k=1}^{\mathrm{\infty }}{e}^{tk}{q}^{(k-1)}p=p{e}^t\sum _{k=0}^{\mathrm{\infty }}{(e^{t}q)}^k.$$

The last expression can be simplified as

$$G(t)=\frac{p{e}^t}{1-e^{t}q}.$$

The first derivative is

$$G^{\prime }(t)=\frac{e^{t}p}{{(1-e^{t}q)}^2}$$

so the mean is

$$\mu =E[X]=G^{\prime }(0)=\frac{1}{p}.$$

In order to find the variance, we use the second derivative and thus

$$E[X^2]=G^{\prime \prime }(0)=\frac{2-p}{p^2}$$

and therefore the variance is

$${\sigma }^2=E[X^2]-E{[X]}^2=G^{\prime \prime }(0)-G^{\prime }{(0)}^2=\frac{q}{p^2}.$$

Title	geometric distribution
Canonical name	GeometricDistribution
Date of creation	2013-03-22 13:03:07
Last modified on	2013-03-22 13:03:07
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	14
Author	Mathprof (13753)
Entry type	Definition
Classification	msc 60E05
Synonym	geometric random variable
Related topic	RandomVariable
Related topic	DensityFunction
Related topic	DistributionFunction
Related topic	Mean
Related topic	Variance
Related topic	BernoulliDistribution
Related topic	ArithmeticMean